Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.9.

23 Nov 2020 9:15 am

Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.9,

Question 1.
Find the sum of the following series
(i) 1 + 2 + 3 + …….. + 60
(ii) 3 + 6 + 9 + …….. +96
(iii) 51 + 52 + 53 + …….. + 92
(iv) 1 + 4 + 9 + 16 + …….. + 225
(v) 62 + 72 + 82 + …….. + 212
(vi) 103 + 113 + 123 + …….. + 203
(vii) 1 + 3 + 5 + …… + 71
Solution:
(i) 1 + 2 + 3 + …….. + 60 = 60×612
[Using n(n+1)2 formula]
= 1830.

(ii) 3 + 6 + 9 + …….. + 96 = 3(1 + 2 + 3 + ……… + 32)
= 3×32×332
= 1584

(iii) 51 + 52 + 53 + …….. + 92 = (1 + 2 + 3 + ……. + 92) – (1 + 2 + 3 + …… + 50)
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.9
= 4278 – 1275
= 3003

(iv) 1 + 4 + 9 + 16 + …….. + 225 = 12 + 22 + 3+ 42 + ………… + 152
15×16×316
[using n(n+1)(2n+1)6] formula
= 1240

(v) 62 + 72 + 82 + …….. + 212 = 1 + 22 + 32 + 42 + ………… + 212 – (1 + 22 + ………… + 52)
= 21×22×436 – 5×6×116
= 3311 – 55
= 3256.

(vi) 103 = 113 + 123 + …….. + 203 = 13 + 23+ 33 + ………… + 203 – (13 + 23 + 33 + …………. + 93)
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.9
[Using (n(n+1)2)2 formula]
= 2102 – 452 = 44100 – 2025
= 42075

(vii) 1 + 3 + 5+ … + 71
Here a = 1; d = 3 – 1 = 2; l = 71


Question 2.
If 1 + 2 + 3 + …. + k = 325 , then find 13 + 23+ 33 + …………. + k3
Answer:
1 + 2 + 3 + …. + k = 325
k(k+1)2 = 325 ……(1)
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.9
= 3252 (From 1)
= 105625.

Question 3.
If 13 + 23 + 33 + ………… + K3 = 44100 then find 1 + 2 + 3 + ……. + k
Answer:
13 + 23 + 33 + ………….. + k3 = 44100
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.9
k(k+1)2 = 44100−−−−−√ = 210
1 + 2 + 3 + …… + k = k(k+1)2
= 210


Question 4.
How many terms of the series 13 + 23 + 33+ …………… should be taken to get the sum 14400?
Answer:
13 + 23 + 33 + ……. + n3 = 14400
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.9
n(n+1)2 = 14400−−−−−√
n(n+1)2 = 120 ⇒ n2 + n = 240
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Ex 2.9 25
n2 + n – 240 = 0
(n + 16) (n – 15) = 0
(n + 16) = 0 or (n – 15) = 0
n = -16 or n = 15 (Negative will be omitted)
∴ The number of terms taken is 15


Question 5.
The sum of the squares of the first n natural numbers is 285, while the sum of their cubes is 2025. Find the value of n.
Answer:
12 + 22 + 32 + …. + n2 = 285


Question 6.
Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm, …, 24 cm. How much area can be decorated with these colour papers?
Answer:
Area of 15 square colour papers
= 102 + 112 + 122 + …. + 242
= (12 + 22 + 32 + …. + 242) – (12 + 22 + 92)
= 24×25×496−9×10×196
= 4 × 25 × 49 – 3 × 5 × 19
= 4900 – 285
= 4615
Area can be decorated is 4615 cm2


Question 7.
Find the sum of the series (23 – 1)+(43 – 33) + (63 – 153) + …….. to
(i) n terms
(ii) 8 terms
Answer:
Sum of the series = (23 – 1) + (43 – 33) + (63 – 153) + …. n terms
= 23 + 43 + 63 + …. n terms – (13 + 33 + 53+ …. n terms) …….(1)
23 + 43 + 63 + …. n = ∑(23 + 43 + 63 + ….(2n)3]
∑ 23 (13 + 23 + 33 + …. n3)
= 8 (n(n+1)2)2
= 2[n (n + 1)]2
13 + 33 + 53 + ……….(2n – 1)3 [sum of first 2n cubes – sum of first n even cubes]

Substituting (2) and (3) in (1)
Sum of the series = 2n2 (n + 1)2 – n2 (2n + 1)2 + 2n2(n + 1)2
= 4n2 (n + 1)2 – n2 (2n + 1)2
= n2 [(4(n + 1)2 – (2n + 1)2]
= n2 [4n2 + 4 + 8n – 4n2 – 1 – 4n]
= n2 [4n + 3]
= 4n3 + 3n2

(ii) when n = 8 = 4(8)3 + 3(8)2
= 4(512) + 3(64)
= 2240.


VISITORS COUNT

166880
Users Today : 248
Total Users : 166879
Views Today : 633
Total views : 617446

Browse Categories

Archives